How to Solve Complex Trigonometry Equations?

[vijay123]

anyway...i didnt mean to start any fight...it maybe some people mis read wut i wrote which caused the whole rpblem...i am truly saying that i got the same answer, just that i thought it wus wrong...since the answer is in a logarithmic form!hence, i thought that it would be impossible to simplify the problem...hence i sought the help of you guys...

 

[vijay123]

if you guys still don't believe me...then go check post no. 7! 7...yes...number SEVEN of this thread. i got the answer bfore, jus that i thought it wus wrong...any more arguments to continue?

 

[VietDao29]

...
240=2^4x/3^2x
hence x=0.5((log240))/(log(4/3))

This is fine, but you can simplify it a little bit further. Using the Change base formula:
\log_{a} b = \frac{\log_c b}{\log_c a}, we have:
x = \frac{1}{2} \times \frac{\log 240}{\log \left( \frac{4}{3} \right)} = \frac{1}{2} \times \log_{\frac{4}{3}} 240.
------------------
Or even faster, given that a^x = y, the unknown x is just x = log_ay
So you can do like this:
240 = \frac{2 ^ {4x}}{3 ^ {2x}}
\Leftrightarrow 240 = \frac{16 ^ {x}}{9 ^ {x}}
\Leftrightarrow 240 = \left( \frac{16}{9} \right) ^ x
\Leftrightarrow x = \log_{\frac{16}{9}} 240
----------------
Is the log in your second problem is in base e or base 10?
And by the way, there's no fight here. You didn't start any. :)

 

[vijay123]

uhh...tis base 10...
thanks for the simplification...

 

[VietDao29]

Ok, let's move on to the second question then:

anyway...for the second answer...i got mine as x=2/9...

How did you get x = 2 / 9? It does not look quite right to me.
It can be done exactly the same as the first problem. So would you mind if you show us how you get x = 2 / 9? Or do you need some help with this one also?
Just give it a try, and see what you get. If you get stuck somewhere, just podt it here.

 

[vijay123]

see, here's how i did it...2^log2/3x^log2=(9x)^log3
((2^log2)/(3^log3))^(1/log6)=3x
2^log(6)2/3^log(6)3=3x
2/3=3x
x=2/9

 

[vijay123]

sry mist some steps...but u ll get the picture once you do it yourself...you think the method is correct?...the ans. is correct coz i checked it in my graphing calc.

 

[VietDao29]

Uhm, sorry for misreading the problem. But by the way, you should use parentheses to make the expressiona clearer. Saying:

b) (2/3x)^(log2) = (9x)^(log3)

is a bit confusing. Since, there are 2 ways to read it.
Is it:
\left( \frac{2}{3} x \right) ^ {\log 2} = (9x) ^ {\log 3}
or:
\left( \frac{2}{3x} \right) ^ {\log 2} = (9x) ^ {\log 3}
-------------
Your work looks good to me. The method is correct. Just missing a few steps. Congratulations. :)
By the way, do you know how to show that:
\frac{2 ^ {\log_6 2}}{3 ^ {\log_6 3}} = \frac{2}{3}?

 

[vijay123]

yes...we have learned how to prove it...thanks for your help

 

[vijay123]

omg...how do you show that 2^(log2/log6)=2, by the case above...
i have only learned that 2^(log6/log2)=6!

 

[VietDao29]

omg...how do you show that 2^(log2/log6)=2, by the case above...

No, this is not correct.
2 ^ {\frac{\log 2}{\log 6}} = 2 ^ {\log_6 2} \neq 2!
---------------
In general, you can show that:
\frac{a ^ {\log_{ab}a}}{b ^ {\log_{ab}b}} = \frac{a}{b}.
Since the base of the log in both numerator and denominator is ab. So, to apply the property of inverse function: f(f^-1(x)) = x, or f^-1(f(x)) = x, i.e (ab) ^ {\log_{ab} x} = x.
We should first multiplying both numerator, and denominator by either b ^ {\log_{ab}a} or a ^ {\log_{ab}b} to make it have the form above.
So:
\frac{a ^ {\log_{ab}a}}{b ^ {\log_{ab}b}} = \frac{\left( a ^ {\log_{ab}a} \right) \left( a ^ {\log_{ab}b} \right)}{\left( b ^ {\log_{ab}b} \right) \left( a ^ {\log_{ab}b} \right)} = \frac{a ^ {\log_{ab} a + \log_{ab} b}}{\left( ab \right) ^ {\log_{ab} b}} = ...
Apply this to your problem, Note that: 6 = 2 . 3
Can you go from here? :)